Wednesday, May 8, 2013

A cute application of Burnside's lemma

Burnside's lemma Let $X$ be a finite set and $G$ a finite group acting on $X$,$$ G\times X\to X,\;\; (g,x)\mapsto g\cdot x. $$Denote by $G\backslash X$ be the space of orbits of this action. For each $g\in G$ we denote by $X^g$ the set of points in $X$ fixed by $g$$$X^g:=\lbrace x\in X;\;\;g\cdot x=x\rbrace. $$Then$$ |G\backslash X|=\frac{1}{|G|} \sum_{g\in G} |X^g|, $$where $|A|$ denotes the cardinality of a set $A$.

For any $x\in X$ the fiber $r^{-1}(x)$ can be identified with the stabilizer $G_x$ of $x$,
$$ G_x=\bigl\lbrace g\in G;\;\; g\cdot x =x\bigr\rbrace. $$
If $x, y$ are in the same orbit of $G$ then $|G_x|=|G_y|$. Indeed, if $y=g\cdot x$ then $G_y=gG_xG^{-1}$.

The function $x\mapsto |r^{-1}(x)|$ is thus constant along the orbits of $G$ on $X$. Thus the contribution of a single orbit $G\cdot x_0$ to the sum $\sum_x|r^{-1}(x)|$ is $|G_{x_0}|\cdot |G\cdot x_0|=|G|$. This shows that

I can now discuss the problem that prompted this post. It is about certain planar lattice walks. The allowable steps of this walk belong to the collection $\newcommand{\eS}{\mathscr{S}}$ $\newcommand{\be}{\boldsymbol{e}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bs}{\boldsymbol{s}}$

We denote by $\newcommand{\eA}{\mathscr{A}}$ $\eA_n$ the set of admissible walks of length $n$. Since $\bs_1$ can be chosen in four different ways, while each of the following steps can be chosen in three different ways so as to preserve (\ref{2}) we deduce

$$ |\eA_n|= 4\cdot 3^{n-1}. \tag{3}\label{3} $$

There is however a symmetry in the story, i.e., a group acting on $\eA_n$. $\newcommand{\eO}{\mathscr{\eO}}$ Denote by $G_0$ the subgroup of $O(2)$ generated by the reflections $R_0, R_1, R_2$ where

$$R_0\be_1=\be_2,\;\;R_0\be_2=\be_1, $$

$$R_1\be_1=\be_1,\;\;R_1\be_2=-\be_2, $$

$$R_2\be_1=-\be_1,\;\;R_2\be_2=-\be_2. $$

This group has order $8$ and can be identified with the Weyl group of $O(4)$.

The group $G_0$ acts on $\bR^2$ and $\eS\subset \bR^2$ is a $G_0$-invariant subset. In particular, $G_0$ acts on the set $\eS^n$ of walks of length $n$, and $\eA_n\subset \eS^n$ is $G_0$-invariant. Observe that if $g\in G_0$ then

There exists another involution $R_3$ that acts on $\eS^n$. More precisely

$$R_3(\bs_1,\dotsc,\bs_n)=(\bs_n,\dotsc,\bs_1). $$

Clearly $R_3\eA_n\subset\eA_n$. We denote by $G$ the subgroup of the group of permutations of $\eA_n$ generated by $G_0$ and $R_3$. Since obviously $R_3$ commutes with all of the reflections $R_0,R_1, R_2$ we deduce that $G\cong G_0\times (\bZ/2). $

We would like to compute the number of orbits of the action of $G$ on $\eA_n$. We denote by $p_n$ this number. The final answer is contained in the equalities (\ref{odd}) and (\ref{even}) below which show different behaviors depending on whether $n$ is odd or even. Here are the details.

For any $\gamma\in G$ we denote by $\eA_n(\gamma)$ the set of admissible paths of length $n$ fixed by $\gamma$ and we set $p_n(\gamma):=|\eA_n(\gamma)|$. From Burnside's lemma we deduce

This shows that the map $G_0\ni g\mapsto |\eA_n(g, R_3)|$ is constant on the conjugacy classes of $G_0$. Let us describe these conjugacy classes. It helps to observe that the morphism $\det :O(2)\to\{\pm 1\}$ induces a morphism

$$ \det :G_0\to\{\pm 1\} $$

and this morphism is constant on the conjugacy classes. There exist two conjugacy classes of determinant $-1$

$$ C(R_1)=\{ R_1, R_2=-R_1\},\;\; C(R_0)=\{R_0, -R_0\}.$$

The other conjugacy classes are

$$ C_1=\{1\},\;\; C_{-1}= \{-1\}, \;\; C=\{ J, -J\}, $$

where $J= RR_1$ is the counterclockwise rotation by $\pi/2$. Thus we need to compute the five numbers